Integrand size = 26, antiderivative size = 205 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d} \]
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Time = 0.59 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3639, 3676, 3673, 3608, 3561, 212} \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rule 212
Rule 3561
Rule 3608
Rule 3639
Rule 3673
Rule 3676
Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^3(c+d x) \left (-4 a+\frac {13}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan ^2(c+d x) \left (-\frac {63 i a^2}{2}-\frac {141}{4} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {267 a^3}{2}-\frac {1083}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {\int \sqrt {a+i a \tan (c+d x)} \left (\frac {1083 i a^3}{8}+\frac {267}{2} a^3 \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {\text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d} \\ \end{align*}
Time = 2.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-15 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-707 i+1760 \tan (c+d x)+1305 i \tan ^2(c+d x)-200 \tan ^3(c+d x)+40 i \tan ^4(c+d x)\right )}{(-i+\tan (c+d x))^3}}{120 a^3 d} \]
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Time = 1.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {31 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}}{a^{4} d}\) | \(131\) |
default | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {31 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}}{a^{4} d}\) | \(131\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (160) = 320\).
Time = 0.25 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.67 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (983 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1527 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 348 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 33 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]
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\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 160 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 1440 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {12 \, {\left (155 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} - 30 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 4 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{6} d} \]
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\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 5.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.68 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^4\,d}-\frac {\frac {31\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {3\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}+\frac {a^2}{5}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}\,d} \]
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